The Poisson distribution with $\lambda=1/2$ is the discrete probability distribution of the number of customers arriving in one minute. It takes values in the set $\{0,1,2,3,\ldots\}$. The time between arrivals of successive customers is a continuous random variable, taking values in $(0,\infty)$.
Let $T$ be that random variable. The probability $\Pr(T>t)$ is the same as the probability that the number of customer arriving before time $t$ is $0$. That number of customers has a Poisson distribution with expected value $\lambda t = t/2$. The probability that that is $0$ is therefore $\dfrac{(t/2)^0 e^{-t/2}}{0!}=e^{-t/2}$. If you plug in $t=2$, you get the answer to your first question.
For your second question, the probability that it's more than one minute you get from plugging in $t=1$, and similarly for the probability that it's more than three minutes. So now you need this:
$$
\Pr(1<T<3) = \Pr(T>1\ \&\ T\not>3).
$$
You need to show this last probability is
$$
\Pr(T>1)-\Pr(T>3).
$$
That's the same as showing
$$
\Pr(T>1\ \&\ T\not>3) + \Pr(T>3) = \Pr(T>1).
$$
So show that the two events $[T>1\ \&\ T\not>3]$ and $[T>3]$ are mutuallly exclusive and that if you put "or" between them, what you get is equivalent to the event $[T>1]$.
We need to decide between minutes and hours for our unit of time. Say it is minutes. Then the mean time between arrivals is $3$ minutes. Or else, depending on the way the exponential distribution has been introduced to you, the rate is $1/3$.
Recall that an exponential distribution with parameter $\lambda$ has mean $\frac{1}{\lambda}$. So $\frac{1}{\lambda}=3$ and therefore $\lambda=\frac{1}{3}$.
A customer has just arrived. Let $X$ be the waiting time until the next customer arrives. Then $X$ has exponential distribution with parameter $\lambda=\frac{1}{3}$. For any positive $x$,
$$\Pr(X\le x)=\int_0^x \frac{1}{3}e^{-t/3}\,dt=1-e^{-x/3}.\tag{$1$}$$
Now we can answer the questions. Interpretation is needed, since there are some ambiguities in the questions.
(a) Interpret the question as saying: "Customer Alicia has just arrived. What is the probability that there will be a customer who arrives later than Alicia, but no more than $3$ minutes later." Then we want $\Pr(X\le 3)$. By $(1)$, this is $1-e^{-3/3}$.
(b) Interpret the question as saying that Alicia has just arrived, and we want the probability that there will be a gap of at least $6$ minutes until the next customer arrives. Then we want $\Pr(X\gt 6$. By $(1)$, this is $1-\Pr(X\le 6)$, which is $1-(1-e^{-6/3})$.
Remark: The exponential distribution is at best a crude model of the situation. For one thing, banks do close. For another, there is always a long line when one is in a hurry.
Best Answer
People arrive at a variety store once every 10 minutes or so.
(a) What is the probability nobody arrives in the next 15 minutes?
Ans: 0.223
(b) What is the probability at least 3 people arrive in the next 20 minutes?
Ans: 0.323
(c) What is the expected number of arrivals in a 10 hour day?
Ans: 60 people
(d) Find the probability the first customer of the day arrives within the first half hour.
Ans: 0.950
(e) Find the probability nobody arrives within the first 30 minutes.
Ans: 0.05
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